Chapter 15, Problem 70SCQ

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# Neither PbCl2 nor PbF2 is appreciably soluble in water. If solid PbCl2 and solid PbF2 are placed in equal amounts of water in separate beakers, in which beaker is the concentration of Pb2+ greater? Equilibrium constants for these solids dissolving in water are as follows:PbCl2(s) ⇄ Pb2+(aq) + 2 Cl−(aq)    Kc = 1.7 × 10−5PbF2(s) ⇄ Pb2+(aq) + 2 F−(aq)    Kc = 3.7 × 10−8

Interpretation Introduction

Interpretation:

The beaker that has higher concentration of Pb+2 has to be identified and explained.

Concept Introduction:

Equilibrium constantKC for a general reaction is:

aA(g)+bB(g)cC(g)+dD(g)KC=[C]c×[D]d[A]a×[B]b

Equilibrium constant in terms of pressure[Kp]: Equilibrium constant can be expressed in terms of partial pressures in atmospheres.

aA(g)+bB(g)cC(g)+dD(g)KP=PCc×PDdPAa×PBb

The activity of solid substance will not appear in equilibrium and numerically it is considered as one.

Explanation

Given:

PbCl2â€‰â€‰(s)â€‰â€‰â€‰â€‰â‡„â€‰Pb+2â€‰â€‰â€‰+â€‰â€‰2â€‰Cl-â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰KCâ€‰â€‰â€‰â€‰â€‰=â€‰â€‰1.7Ã—10-5PbF2â€‰(s)â€‰â€‰â‡„â€‰â€‰Pb+2â€‰â€‰â€‰+â€‰â€‰2â€‰F-â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰KCâ€‰â€‰â€‰â€‰â€‰=â€‰â€‰3.7Ã—10-8

Equilibrium constantKC for a general reaction is:

aAâ€‰(g)â€‰+â€‰bB(g)â‡„â€‰cCâ€‰(g)â€‰+â€‰dDâ€‰(g)â€‰â€‰KCâ€‰â€‰â€‰â€‰â€‰â€‰=â€‰â€‰[C]câ€‰â€‰Ã—â€‰[D]d[A]aâ€‰Ã—â€‰[B]b

In the given reaction PbCl2 and PbF2 are in solid state and thus its concentration can be taken as unity. There is no role of PbCl2 and PbF2 in the equilibrium expression.

Only the concentration of Pb+2â€‰,â€‰Cl-â€‰,â€‰F- appears in the equilibrium constant expression for the given reactions.

PbCl2â€‰â€‰(s)â€‰â€‰â€‰â€‰â‡„â€‰Pb+2â€‰â€‰â€‰+â€‰â€‰2â€‰Cl-â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰KCâ€‰â€‰â€‰â€‰â€‰=â€‰â€‰1

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